Question

How do you determine `(d^2y)/(dx^2)` given `x^(2/3)+y^(2/3)=8`?

Answer 1

Implicit differentiation, stubbornness and attention to details.

Explanation:

The more challenging (difficult) a problem is, the MORE you should write. (As a student I tried to skip steps and it took me a year of two to learn this important fact. If you can answer in two lines, do it in one. If you think it will take 8 lines, plan to write 12.)

`x^(2/3)+y^(2/3) = 8`

Differentiate

`2/3x^(-1/3) + 2/3y^(-1/3) dy/dx =0`

Multiply through b `3/2` to get rid of those fractions.

`x^(-1/3) + y^(-1/3) dy/dx =0`.

Solve for `dy/dx`

`y^(-1/3) dy/dx = - x^(-1/3)`

`dy/dx = -x^(-1/3)/y^(-1/3)`

It's usually silly to have a negative exponent in a fraction, so let's fix that.

`dy/dx = - y^(1/3)/x^(1/3)`

Now differentiate again.

`(d^2y)/dx^2 = ((-1/3 y^ (-2/3) dy/dx)(x^(1/3))-(-y^(1/3))(1/3x^(-2/3)))/x^(2/3)`

`= (- 1/3(x^(1/3)/y^(2/3))dy/dx + 1/3(y^(1/3)/x^(2/3)))/x^(2/3)`

Let's factor out that `1/3` (We could take a `-1/3`, but `dy/dx` has a minus in front of it.)

`=1/3 (- (x^(1/3)/y^(2/3))dy/dx + (y^(1/3)/x^(2/3)))/x^(2/3)`

Use `dy/dx = - y^(1/3)/x^(1/3)` to get

`= 1/3 (- (x^(1/3)/y^(2/3))(- y^(1/3)/x^(1/3)) + (y^(1/3)/x^(2/3)))/x^(2/3)`

`= 1/3 ( (1/y^(1/3)) + (y^(1/3)/x^(2/3)))/x^(2/3)`

Clear the denominators in the numerator by multiplying the big fraction by `((x^(2/3)y^(1/3)))/((x^(2/3)y^(1/3)))`

`= 1/3 (((1/y^(1/3)) + (y^(1/3)/x^(2/3))))/(x^(2/3)) * ((x^(2/3)y^(1/3)))/((x^(2/3)y^(1/3)))`

`= 1/3 (x^(2/3)+y^(2/3))/( x^(4/3)y^(1/3))`

Finally, notice that the numerator is the expression we started with and is equal to `8`.

`(d^2y)/dx^2 = 8/(3x^(4/3)y^(1/3))`

Answer 2

`y'' =1/3y^(1/3)x^(-4/3)-1/3x^(-1/3)y^(-2/3)y'`

Explanation:

Differentiate the equation with respect to `x`:

`d/(dx) (x^(2/3) + y^(2/3)) = 0`

`2/3x^(-1/3) + 2/3y^(-1/3) y' = 0`

Solving for `y'` we have:

`y'=-x^(-1/3)/y^(-1/3) = -y^(1/3)/x^(1/3)`

Differentiating again:

`y'' = -(1/3x^(1/3)y^(-2/3)y'-1/3y^(1/3)x^(-2/3))/x^(2/3)=1/3y^(1/3)x^(-4/3)-1/3x^(-1/3)y^(-2/3)y'`